【JAVA、C++】LeetCode 004 Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

解题思路：

由于要求时间复杂度O(log (m+n))所以几乎可以肯定是递归和分治的思想。

《算法导论》里有找两个数组第K小数的算法，时间复杂度为O(log(m+n))，所以直接调用即可

参考链接：http://blog.csdn.net/yutianzuijin/article/details/11499917/

Java参考代码：

public class Solution {
public static double findKth(int[] nums1, int index1, int[] nums2,
            int index2, int k) {
        if (nums1.length - index1 > nums2.length - index2)
            return findKth(nums2, index2, nums1, index1, k);
        if (nums1.length - index1 == 0)
            return nums2[index2 + k - 1];
        if (k == 1)
            return Math.min(nums1[index1], nums2[index2]);
        int p1 = Math.min(k / 2, nums1.length - index1), p2 = k - p1;
        if (nums1[index1 + p1 - 1] < nums2[index2 + p2 - 1])
            return findKth(nums1, index1 + p1, nums2, index2, k - p1);
        else if (nums1[index1 + p1 - 1] > nums2[index2 + p2 - 1])
            return findKth(nums1, index1, nums2, index2 + p2, k - p2);
        else
            return nums1[index1 + p1 - 1];
    }

    static public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if ((nums1.length + nums2.length) % 2 != 0)
            return findKth(nums1, 0, nums2, 0,
                    (nums1.length + nums2.length) / 2 + 1);
        else
            return findKth(nums1, 0, nums2, 0,
                    (nums1.length + nums2.length) / 2)
                    / 2.0
                    + findKth(nums1, 0, nums2, 0,
                            (nums1.length + nums2.length) / 2 + 1) / 2.0;
    }
}

C++实现如下：

 #include<vector>
 #include<algorithm>
 using namespace std;
 class Solution {
 private:
     double findKth(vector<int> nums1, int index1, vector<int> nums2, int index2, int k) {
         if (nums1.size() - index1 > nums2.size() - index2) {
             swap(nums1, nums2);
             swap(index1,index2);
         }
         if (nums1.size() - index1 == )
             return nums2[index2 + k - ];
         if (k == )
             return min(nums1[index1], nums2[index2]);
         int p1 = min(k / , (int)nums1.size() - index1), p2 = k - p1;
         if (nums1[index1 + p1 - ] < nums2[index2 + p2 - ])
             return findKth(nums1, index1 + p1, nums2, index2, k - p1);
         else if (nums1[index1 + p1 - ] > nums2[index2 + p2 - ])
             return findKth(nums1, index1, nums2, index2 + p2, k - p2);
         else
             return nums1[index1 + p1 - ];
     }

 public:
     double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
         if ((nums1.size() + nums2.size()) &)
             return findKth(nums1, , nums2, ,
                 (nums1.size() + nums2.size()) /  + );
         else
             return findKth(nums1, , nums2, ,(nums1.size() + nums2.size()) / )/ 2.0
             + findKth(nums1, , nums2, ,(nums1.size() + nums2.size()) /  + ) / 2.0;
     }
 };